# Capacitor Conceptual Question!

## Homework Statement

If you increase the charge on a parallel plate capacitor from 3μC to 9μC and increase the plate separation from 1mm to 3mm, the energy stored in the capacitor changes by a factor of...

U = 1/2QV
or 1/2CV^2

## The Attempt at a Solution

I'm not sure which equation to even use, i'm not to great when it comes to differentiating between electric field and energy of the capacitor.

I think I should probably use 1/2QV but i don't have a voltage to work with so i'm confused.

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Use the relation between charge Q, Voltage V, and Capacitance C to eliminate the voltage from the expression for the energy.

Use the relation between charge Q, Voltage V, and Capacitance C to eliminate the voltage from the expression for the energy.
so U =1/2 Q^2/C

i'm still not sure how i would find capacitance without using the C=Q/V and not having a voltage?
also does the distance between the plates changing not matter?

mfb
Mentor
The capacitance changes if you change the distance, and you should know a formula that connects the two.

The capacitance changes if you change the distance, and you should know a formula that connects the two.
oh alright i thought that the permitivity of free space had to be used when a dielectric was present only but that works thanks!